One of the most effective recruiting techniques to the Virginia Tech Honors program is getting the students to visit the campus before making their decision. According to recent data, 45% of the potential honors students visit the campus before making their decision. If the student visits, they accept the offer to attend Virginia Tech 95% of the time, but the students that do not visit the campus before making their decision will attend 55% of the time. Knowing a student will enter the Virginia Tech Honors program, what is the probability that they visited the campus before making their decision

Wow a real Bayes Theorem question.  

Let's name some events.

A - honor student accepts offer to attend VT

V - honor student visits VT

45% of the potential honors students visit the campus before making their decision

P(V) = .45

If the student visits, they accept the offer to attend Virginia Tech 95% of the time

P(A | V) = 0.95

but the students that do not visit the campus before making their decision will attend 55% of the time.

P(A | ¬V) = 0.55

Knowing a student will enter the Virginia Tech Honors program, what is the probability that they visited the campus before making their decision

We're asked for P(V | A)

OK, according to Bayes Theorem

P(V | A) = P(A | V) P(V) / (  P(A | V) P(V) +  P(A | ¬V) P(¬V) )

Of course P(¬V) = 1 - P(V) = 1 - .45  = .55

I think we have everything to do the calculation.  

P(V | A) = (.95 ×  .45)/ (   .95 ×  .45  + .55 × .55 ) = 0.5856164

58.6%

answer: (1,3) and (-2,0) i think.